Testing Real or Complex Roots: y=g(x)D^{k}f(x)

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    Complex Roots Testing
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SUMMARY

The discussion focuses on determining whether the roots of the function y=f(x) are real or complex, particularly in the context of y=g(x)D^{k}f(x) where k>0 and D=d/dx. It highlights that functions like y=exp(2πx)-1 have only real roots, while y=exp(x^2)+1 has only complex roots despite being real for all x. A definitive test for identifying real roots is to check if the function is always positive or negative, as demonstrated with y=exp(x^2)+1, which is always greater than or equal to 2, confirming it has no real roots.

PREREQUISITES
  • Understanding of real and complex roots in functions
  • Familiarity with exponential functions, specifically y=exp(x)
  • Basic knowledge of calculus, particularly differentiation (D=d/dx)
  • Concept of positivity and negativity in mathematical functions
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  • Research the properties of exponential functions, particularly y=exp(x^2)
  • Study methods for determining the nature of roots in polynomial and non-polynomial functions
  • Explore advanced calculus techniques for analyzing function behavior
  • Learn about the implications of the Intermediate Value Theorem on root existence
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Mathematicians, calculus students, and anyone interested in the analysis of functions and their roots, particularly in advanced mathematics and theoretical physics.

Karlisbad
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For the function ## y=f(x) ## is there a test to prove if its roots are real or either has some complex roots?, or in more general cases:

## y=g(x)D^{k}f(x) ## k>0 and a real D=d/dx number.:rolleyes: :rolleyes:

The question is that sometimes it can be very deceiving to tell if a function has real or complex roots, for example:

## y=exp(2 \pi x)-1 ## has only real roots.. but for real x the function ## y=exp(x^2)+1 ## has only complex roots , but for every real x the function is real.
 
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Karlisbad said:
For the function ## y=f(x) ## is there a test to prove if its roots are real or either has some complex roots?, or in more general cases:

## y=g(x)D^{k}f(x) ## k>0 and a real D=d/dx number.:rolleyes: :rolleyes:

The question is that sometimes it can be very deceiving to tell if a function has real or complex roots, for example:

## y=exp(2 \pi x)-1 ## has only real roots.. but for real x the function ## y=exp(x^2)+1 ## has only complex roots , but for every real x the function is real.
A simple test is whether the function is always positive or is always negative. In your second example, ##y = \exp(x^2) + 1##, ##x^2 \ge 0## for all real x, and ##\exp(x^2) \ge 1## for all real x, so adding 1 makes ##\exp(x^2) + 1 \ge 2##. Therefore, this function can't have any real roots.
 

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